Simplify the following expression and state the condition under which the simplification is valid. $p = \dfrac{2q^2 - 14q - 60}{-6q^3 - 60q^2 - 126q}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ p = \dfrac {2(q^2 - 7q - 30)} {-6q(q^2 + 10q + 21)} $ $ p = -\dfrac{2}{6q} \cdot \dfrac{q^2 - 7q - 30}{q^2 + 10q + 21} $ Simplify: $ p = - \dfrac{1}{3q} \cdot \dfrac{q^2 - 7q - 30}{q^2 + 10q + 21}$ Next factor the numerator and denominator. $ p = - \dfrac{1}{3q} \cdot \dfrac{(q + 3)(q - 10)}{(q + 3)(q + 7)}$ Assuming $q \neq -3$ , we can cancel the $q + 3$ $ p = - \dfrac{1}{3q} \cdot \dfrac{q - 10}{q + 7}$ Therefore: $ p = \dfrac{ -q + 10 }{ 3q(q + 7)}$, $q \neq -3$